Sedimentation (Under a Gravitational Potential)

Chemistry 24b Lecture 6 Spring Quarter 2003 Date: April 14 Instructor: Richard Roberts

in fluid

Figure 4-1

dv driving force = fv = m (force = ma) dt

In gravitational potential, = = − ρ = − ρ Fd driving force mg mv 2 g m()1 v2 g where g = gravitational acceleration m = mass of particle f = frictional coefficient = v2 partial specific volume of particle  ∂V  =    ∂  g2 ρ T1 1g1 ρ = density of viscous medium v = velocity of sedimentation

Note that

≡ mv2 effective volume of particle ρ = so that mv2 g buoyant force

Need to include buoyant force! Chemistry 24b — Lecture 6 2

Therefore,

()− ρ − = dv m 1 v2 g fv m dt dv When the frictional force balances the driving force, = 0 and the particle acquires its () dt terminal velocity vt .

− ρ m()1 v2 g v = t f

Sedimentation (in an Ultracentrifuge) The same principles and formulae apply except that g is replaced by ω 2x . ()− ρ ω 2 = m 1 v 2 x vt f where ω ≡ speed of rotation (in radians / sec) x ≡ distance from the axis of rotation

axis of rotation meniscus

cup x

Figure 4-2

Sedimentation Coefficient (S)

v m()1 − v ρ S = t = 2 ω 2x f where S ≡ terminal velocity / unit acceleration

Sedimentation coefficients have units of sec. 10−13 sec is called 1 svedberg (or 1 S). T. Svedberg pioneered research on sedimentation in an ultracentrifuge.

−13 1 S ≡ 10 sec Chemistry 24b — Lecture 6 3

Determination of Sedimentation Coefficient

(1) Boundary Sedimentation Boundary sedimentation is sedimentation of macromolecules in a homogeneous solution. (a) Begin with a homogeneous solution of macromolecules. (b) As the solution is spun in the ultracentrifuge and macromolecules move down the centrifugal field, a solution-solvent boundary is generated. The boundary can be monitored by refractive index, color (absorption) etc.

time 0 C0 plateau region

Concentration time t solution solvent 0 x1/2 xm xb x meniscus cell bottom

Figure 4-3

v dx / dt 1 d ln x S = t = 1/2 = 1/ 2 ω 2 ω 2 ω 2 x x1/2 dt slope

2.303 d log x1/ 2 = logx1/2 ω 2 dt 2.303 = (slope) ω 2 t

Figure 4-4 Chemistry 24b — Lecture 6 4

(2) Zone Sedimentation (a) A thin layer of a solution of the macromolecule(s) is placed on top of a solvent

containing sucrose (sucrose solution). band containing solution of macromolecules

buffer + sucrose

Figure 4-5

(b) As the sample is spun in the centrifuge, a band containing macromolecules will move down the centrifugal field. Also, a sucrose gradient will have developed. The sucrose gradient ensures that the density of the “solvent” is always greater than the density of the sedimenting zone. This ensures the stability of the band.

meniscus

sucrose density gradient

species 1 species 2

Figure 4-6

Frictional Coefficient Once S is determined, f can be determined from

m()1− v ρ S = 2 f Chemistry 24b — Lecture 6 5

ρ since m, v 2, and can be determined experimentally.

Sedimentation-Equilibrium (1) Consider sedimentation of a homogeneous two-component solution (a solute plus solvent) in an ultracentrifuge. Because of sedimentation, a concentration gradient is generated. Diffusion sets in. Since transport by sedimentation and diffusion go in opposite directions, eventually an equilibrium concentration is generated by ultracentrifugation. This occurs when k T dC B = m ()1− v ρ ω 2x or C dx 2 RT dC = M ()1− v ρ ω 2x or C dx 2 4.606 RT d log C M = ω 2 − ρ 2 ()1 v2 dx

Measure log C vs x2 to obtain M! (2) Another way to reach sedimentation equilibrium is when the macromolecule becomes “buoyant” in a density gradient. Under this condition,

()− ρ = = 1 v 2 (x) 0 and vt or S 0

(a) In these experiments, a density gradient is established by adding concentrated salt solution, e.g. CsCl, to the solution of macromolecule(s). (b) The solution is then spun at high speeds (ω). (c) The various macromolecular species will form bands at points in the salt gradient where the macromolcules become buoyant; i.e., at x’s where ()− ρ = 1 v2 (x) 0 for the species. Many biological macromolecules have “buoyant densities” sufficiently different that they can be separated or resolved by density-gradient centrifugation. Chemistry 24b — Lecture 6 6

Figure 4-7 Chemistry 24b — Lecture 6 7

Fick's Second Law

∆ Jk (x) Jk (x + x)

x x + ∆x

Figure 4-8

≠ +∆ If Jk (x) Jk(x x) , there will be an accumulation or depletion of the molecular species k from the zone. From mass balance, − +∆ − ∆ = ∆ ⋅ ⋅∆ ⋅∆ = ∆ []Jk(x x) Jk (x) t Ck 1 x , 1 x V

∆ Ck is in moles / cc ∆J x ∆C − k = k ∆x ∆t dC (x) But J x = − D k k dx

Therefore d2C (x,t) ∂C (x,t) D k = k Fick' s Second Law of Diffusion dx2 ∂t ∂ v ∇2 v = Ck(r ,t) or D Ck(r ,t) in 3- dimensions isotropic medium ∂t

Solution to differential equation for a plane source: (a) 1-D only. (b) As t → 0, C(x,0) = 0 for all x' s except at x = 0, where C(0,0) →∞.

(c) Solution 2 −  x  C(x,t) = α t 1/ 2 exp −   4Dt where α ≡ a constant (d) To obtain α, use normalization condition, i.e., conservation of molecules of that species. Chemistry 24b — Lecture 6 8

∞ N = number of molecules = ∫ C(x,t)dx −∞ ∞ 2 −  x  = α t 1/ 2 exp −  dx = 2α(πD)1/2 or ∫   −∞ 4Dt N  x2  C(x,t) = exp −  Gaussian in x! 2(πDt)1/2  4Dt

1 (e) Pr obability of finding molecules at position x at time t ≡ P(x,t) ≡ C(x,t) N This result can be used to calculate mean-square displacement of the molecules from the origin

∞ x 2 = ∫ x 2P(x,t)dx = 2Dt −∞ x 2 = 2Dt

(f)

t = 0

> > t1 t2 t1 0

−x x = 0 x

Figure 4-9 Chemistry 24b — Lecture 6 9

How to Measure D?

(a)

cover slip with

cover slip

+ thin film of solvent

thin strip of colored protein solution Carefully lay the cover slips on top of one another solution face to solution

face and squeeze cover slips

Measure C(x) as a function of time and compare observed concentration profiles with simulated curves to obtain D

Figure 4-10

(b) In 2-D, replace the plane source by a circular patch.

cover slip with

cover slip

+ bilayer membrane film

fluorescence labelled

or spin-labelled

phospholipids

2 2

< x +y > = 4Dt

Figure 4-11 Chemistry 24b — Lecture 6 10

The lateral diffusion coefficient of lipids in bilayer membranes is determined in this fashion (D ~ 10-8 cm2/sec).

− x 2 + y2 = 4Dt = 4 ×10 8 cm2 / sec ()t − r2 = 4 ×10 8 cm2 in 1 sec

1/ 2 − r 2 = 2 ×10 4 cm in 1 sec = µ µ = 2 ( microns)

(c) Quasi-Electric Light Scattering ν Dephasing of a coherent laser beam of light frequency 0 due to Doppler broadening of scattered light.

I0 monochromatic Iscattered

Figure 4-12 Chemistry 24b — Lecture 6 11

Intensity of Scattered Light

Incident light

α 4 D ~10 Hz for protein molecules with − 6 2 D ~10 cm /sec

ν 15 0 ~ 10 Hz

Figure 4-13

Will elaborate on later.